# Question #d8c45

Jul 27, 2016

Given

$\cos \theta + \sin \theta = \sqrt{2} \cos \theta$

Following the formula

${\left(a + b\right)}^{2} + {\left(a - b\right)}^{2} = 2 \left({a}^{2} + {b}^{2}\right)$

we can wite

${\left(\cos \theta + \sin \theta\right)}^{2} + {\left(\cos \theta - \sin \theta\right)}^{2} = 2 \left({\cos}^{2} \theta + {\sin}^{2} \theta\right)$

Putting

$\cos \theta + \sin \theta = \sqrt{2} \cos \theta \text{ }$ we get

${\left(\sqrt{2} \cos \theta\right)}^{2} + {\left(\cos \theta - \sin \theta\right)}^{2} = 2 \cdot 1$

$\implies 2 {\cos}^{2} \theta + {\left(\cos \theta - \sin \theta\right)}^{2} = 2 \cdot 1$

$\implies {\left(\cos \theta - \sin \theta\right)}^{2} = 2 - 2 {\cos}^{2} \theta$

$\implies {\left(\cos \theta - \sin \theta\right)}^{2} = 2 \left(1 - {\cos}^{2} \theta\right)$

$\implies {\left(\cos \theta - \sin \theta\right)}^{2} = 2 {\sin}^{2} \theta$

$\implies \cos \theta - \sin \theta = \sqrt{2} \sin \theta$

Proved

In other way when
$\cos \theta - \sin \theta = \sqrt{2} \sin \theta \text{ }$ we get

${\left(\cos \theta + \sin \theta\right)}^{2} + {\left(\sqrt{2} \sin \theta\right)}^{2} = 2 \cdot 1$

$\implies {\left(\cos \theta + \sin \theta\right)}^{2} + 2 {\sin}^{2} \theta = 2$

$\implies {\left(\cos \theta + \sin \theta\right)}^{2} = 2 - 2 {\sin}^{2} \theta$

$\implies {\left(\cos \theta + \sin \theta\right)}^{2} = 2 \left(1 - {\sin}^{2} \theta\right)$

$\implies {\left(\cos \theta + \sin \theta\right)}^{2} = 2 {\cos}^{2} \theta$

$\implies \cos \theta + \sin \theta = \sqrt{2} \cos \theta$

Proved